What are the asymptote(s) and hole(s), if any, of f(x)= (3e^(x))/(2-2e^(x))?

Feb 22, 2018

There is a horizontal asymptote at $y = - \frac{3}{2}$ and a vertical asymptote at $x = 0$. There are no holes in this function.

Explanation:

To understand how to find asymptotes, you have to understand the definition of asymptotes.

The definition of a vertical asymptote is ${\lim}_{x \rightarrow c} f \left(x\right) = \pm \infty$ with c being the vertical asymptote. In other words, there is a vertical asymptote at $x = c$ when the function approaches infinity or negative infinity. For a rational function like this one, you simply just find an x that makes the denominator zero since that would make the function approach $\pm \infty$.

$2 - 2 {e}^{x} = 0$
$2 {e}^{x} = 2$
${e}^{x} = 1$
$x = \ln 1$
$x = 0$

There is a vertical asymptote at $x = 0$

The definition of a horizontal asymptote is ${\lim}_{x \rightarrow \pm \infty} f \left(x\right) = c$. In other words, there is a horizontal asymptote if, when the x values of the function approach positive or negative infinity, the function converges to a constant. The horizontal asymptote is $f \left(x\right) = c$.

One way to do this is by calculus using the L'Hospital's rule, but that is a bit too long to explain. An easier way is to just use sketchy math- both will give you the same answer in most cases.

To solve using sketchy math:

One assumption we can make is that 2 is negligible because infinity is so large so we just ignore it or set it equal to zero. (I know, not the best way to do math)

$\frac{3 {e}^{x}}{- 2 {e}^{x}}$

As you can see, you can cancel out the ${e}^{x}$'s resulting in

$- \frac{3}{2}$

So as f(x) goes to infinity or negative infinity, the value of the function will go to $- \frac{3}{2}$.

There is a horizontal asymptote at $y = - \frac{3}{2}$. ($f \left(x\right)$ and y are interchangable in this case)

There are no holes because you can't cancel anything out from the top and bottom.

Here is the solution for the horizontal asymptote using L'Hopitals rule if you are curious: